Suppose there are k old ladies ahead of you in line and j people from
yourself to the end of the line (inclusive).  Then the probability that
you get your own seat is j/(j+k).

Proof by induction over m, the number of people ahead of you in line.  

Base case:  m=0.  With nobody ahead of you in line, clearly you're sure to
get your own seat.  And in this case necessarily k=0, so j/(j+k) = j/j = 1.

Inductive case:  Suppose the answer is j/(j+k) when there are fewer than m
people ahead of you in line, and suppose now that there are m people ahead
of you.

Subcase 1:  If the first person in line is NOT an old lady, then that
person simply sits down in his or her own seat and the chance of you
getting your seat doesn't change.  The number of people ahead of you is
reduced to m-1, but k and j are unchanged.  By induction, the chance that
you'll get your own seat is j/(j+k).

Subcase 2:  If the first person in line IS an old lady, then there are
five things that can happen.  Let n be the total number of people in
the line.

   a) With probability 1/n, the old lady takes your seat, and you're SOL.
      You get your seat with probability 0.

   b) With probability 1/n, the old lady takes her own seat.  This reduces
      to the case of m-1 people in front of you, k-1 of which are old ladies,
      so by the inductive hypothesis the chance that you get your own seat
      is j/(j+k-1).

   c) With probability (k-1)/n, the old lady takes the seat of some other
      old lady in front of you.  In this case, it's exactly as if the old
      lady had taken her own seat.  (Old ladies are interchangeable since
      they ignore what's on their ticket anyway---we might as well have
      swapped the tickets of these two old ladies in the first place.) So
      again the chance that you get your own seat is j/(j+k-1).
      
   d) With probability (n-j-k)/n, the old lady takes the seat of some
      NON-old lady NOL in front of you.  The effect here is to transform
      NOL into an old lady!  (After all, NOL will now ignore the seat on
      his or her ticket and pick a seat at random, just as an old lady
      would.)  So now there are m-1 people in front of you, but still k old
      ladies, and the chance that you get your seat is j/(j+k).

   e) With probability (j-1)/n, the old lady takes the seat of someone
      behind you in the line (who may or may not be an old lady).  Once
      again we're reduced to a case with m-1 people in front of you, k-1 of
      which are old ladies, so the chance you get your seat is j/(j+k-1).
      
Multiplying and adding, the chance that you get your seat is

      (1/n)(0) + (1/n)(j/(j+k-1)) + ((k-1)/n)(j/(j+k-1))
               + ((n-j-k)/n)(j/(j+k)) + ((j-1)/n)(j/(j+k-1))

This expression simplifies to j/(j+k), and we're done.